As you noted in the question, if $x +1$ is the first number in the sum, then the sum of the $n$ consecutive numbers is
$$ \frac{n}{2} \cdot (x + 1 + x + n). $$
If $n$ is odd, let $n = 2k + 1$, and take $x = k$. Then the sum of the $n$ consecutive numbers $(k + 1), (k + 2), \dots, (k + n)$ is
$$
\frac{2k+1}{2} \cdot (k + 1 + k + 2k + 1) = (2k + 1)^2
$$
which is a square, so all odd numbers work.
Now suppose that $n$ is even. Let $n = 2k$ for some $k$.
Then the sum is equal to
$$ k(2x + 2k + 1) $$.
The term in brackets is odd, so for this to be a square, the largest power of $2$ which divides $k$ must be even. i.e. We must have that $k = 2^{2m} \cdot d$ where $m$ is some natural number, and $d$ is an odd natural number.
The sum then reduces to
$$ 2^{2m} \cdot d \cdot (2x + 2^{2m + 1} \cdot d + 1) $$
This is a square if and only if
$$ d \cdot (2x + 2^{2m + 1} \cdot d + 1) $$
is a square, so it is enough to choose $x$ so that the term in brackets is a square multiple of $d$. That is, we try to choose $x$ so that
$$ 2x + 2^{2m + 1} \cdot d + 1 = a^2 \cdot d $$
for some natural number $a$.
This is equivalent to
$$ 2x + 1 = \left( a^2 - 2^{2m + 1} \right) \cdot d $$
and it is easy to see that this has a solution for x as long as $a$ is an odd square greater than $2^{2m + 1}$. Thus we have a solution in this case since there are arbitrarily large odd square numbers.
We see that if $n$ is even, then $n$ works if and only if $n$ is divisible by $2$ an odd number of times.